Question

Find radius of curvature of y=e^x at (0,1)​

Answer 1

$\large\underline{\sf{Solution-}}$

Given curve is

$\rm :\longmapsto\:y = {e}^{x}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} y = \dfrac{d}{dx}{e}^{x}$

$\rm :\longmapsto\:y_1 = {e}^{x}$

Again differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y_1 = \dfrac{d}{dx} {e}^{x}$

$\rm :\longmapsto\:y_2 = {e}^{x}$

Now, we calculate the value of differential coefficients at ( 0, 1 ).

$\rm :\longmapsto\:y_1 \: at \: (0,1) = {e}^{0} = 1$

$\rm :\longmapsto\:y_2 \: at \: (0,1) = {e}^{0} = 1$

We know,

Radius of curvature for cartesian curve, y = f(x) is

$\boxed{ \bf{ \: \rho \: = \: \dfrac{ {\bigg[1 + {y_1}^{2} \bigg]}^{ \dfrac{3}{2} } }{ |y_2| } }}$

So, on substituting the values, we get

$\rm :\longmapsto\:{ \bf{ \: \rho \: = \: \dfrac{ {\bigg[1 + {1}^{2} \bigg]}^{ \dfrac{3}{2} } }{ 1 } }}$

$\rm :\longmapsto\: \rho \: = \: {\bigg[1 + 1\bigg]}^{\dfrac{3}{2} }$

$\rm :\longmapsto\: \rho \: = \: {\bigg[2\bigg]}^{\dfrac{3}{2} }$

$\rm :\longmapsto\: \rho \: = \: {\bigg[ {( \sqrt{2} )}^{2} \bigg]}^{\dfrac{3}{2} }$

$\rm :\longmapsto\: \rho \: = \: {\bigg[ {( \sqrt{2} )} \bigg]}^{3}$

$\bf\implies \: \rho \: = \: 2 \sqrt{2}$

Additional Information :-

1. Radius of curvature for polar curves

$\boxed{ \bf{ \: \rho \: = \: \dfrac{ {\bigg[1 + {r_1}^{2} \bigg]}^{ \dfrac{3}{2} } }{ | {r}^{2} +2 {r_1}^{2} - r r_2| } }}$

2. Radius of curvature for parametric curves

$\boxed{ \bf{ \: \rho \: = \: \dfrac{ {\bigg[ {x_1}^{2} + {y_1}^{2} \bigg]}^{ \dfrac{3}{2} } }{ \: \: \: \: | x_1y_2 - y_1x_2| \: \: \: \: \: \: }}}$